Solve: 3sin⁡2x+4cos⁡2x−2cos⁡x+6sin⁡x−6=0 My Try 6sin⁡xcos⁡x+4(cos2x−sin2x)−2cos⁡x+6sin⁡x−6=0

${\displaystyle 6\sin x\cos x+4({\cos }^{2}x-{\sin }^{2}x)-2\cos x+6\sin x-6=0}$
${\displaystyle 6\sin x\cos x-9{\sin }^{2}x-{\cos }^{2}x-2\cos x+6\sin x-1=0}$
${\displaystyle -{(3\sin x-\cos x)}^2+2(3\sin x-\cos x)-1=0}$
${\displaystyle {(3\sin x-\cos x)}^2-2(3\sin x-\cos x)+1=0}$

x=2y we get after some manipulation
${\displaystyle -4{\cos }^{4}y-36{\cos }^{2}y{\sin }^{2}y+24{\cos }^{3}y\sin y=0}$
${\displaystyle -4{\cos }^{2}y{(\cos y-3\sin y)}^2=0}$

Abbreviating $\sin x$ and $\cos x$ to s and c, and writing $\sin 2x=2sc$ and $\cos 2x=2c^2-1$, the equation becomes
$6sc+4(2c^2-1)-2c+6s-6=0$
Factoring out a 2 and grouping terms, we rewrite this as
$3s(c+1)+4c^2-c-5=0$
But $4c^2-c-5=(4c-5)(c+1)$, so we have two solutions in s and c:
c=-1 and 3s+4c=5
Reverting to $\sin x$ and $\cos x$ and recognizing the Pythagorean triple $3^2+4^2=5^2$, these become
$\cos x=-1$ and $\sin (x+\theta )=1$
where $\sin \theta =\frac{4}{5}$ and $\cos \theta =\frac{3}{5}$, e.g. $\theta =\arcsin (\frac{4}{5})$. So the solution set is
$\{\pi +2n\pi |n\in \mathbb{Z}\}\cup \{\frac{\pi }{2}-\arcsin (\frac{4}{5})+2n\pi |n\in \mathbb{Z}\}$